Yes, you are right. You did the algebra, so I'm not sure what more confirmation you need. But think about it conceptually: in a resistor that obeys ohm's law, there is a linear relationship between voltage and current. Hence, if you double the voltage supplied to the resistor, you double the current drawn by it. In other words, this is a question you can do by inspection, too. Same with the power. You should know that the power dissipated in a resistor is given by:

P = V2/R

or

P = I2R

(Do you now how I got these formulae? Eliminate either V or I from P = IV by substituting one expressed in terms of the other.)

..and so this is not a linear, but a quadratic relationship. Doubling either V or I will therefore quadruple the power dissipated. Those formulae allow you to arrive at the answer more directly, but skips out an intermediate step.

However, the way you arrived at the answer is very instructive and useful too! You thought it through step by step: "If I double the voltage, I also end up doubling the current (as we just showed), and so the power quadruples, since it is the product of both."

You can see that there is no difference between the two methods really. Both make use of the relationship between current and voltage, one just does so a little less directly.